Q. \(\displaystyle \frac{d}{dx}\left(\tan^{-1}(x)\right)\)

Q: What is the derivative of \tan^{-1}(x) ?

\dfrac{d}{dx}\left(\tan^{-1}(x)\right)=\dfrac{1}{1+x^2} .

Q: What is the derivative of \tan^{-1}(u) where u depends on x?

Use the chain rule: \dfrac{d}{dx}\left(\tan^{-1}(u)\right)=\dfrac{u'}{1+u^2} .

Q: Find \dfrac{d}{dx}\left(\tan^{-1}(3x)\right) .

Let u=3x, so u'=3. Then \dfrac{d}{dx}\left(\tan^{-1}(3x)\right)=\dfrac{3}{1+(3x)^2}=\dfrac{3}{1+9x^2} .

Q: Find \dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{x-1}{x+1}\right)\right) .

Let u=\dfrac{x-1}{x+1}. Then u'=\dfrac{2}{(x+1)^2}. So derivative is \dfrac{u'}{1+u^2}=\dfrac{\frac{2}{(x+1)^2}}{1+\left(\frac{x-1}{x+1}\right)^2} .

Q: How do you derive \dfrac{d}{dx}\left(\tan^{-1}(x)\right)=\dfrac{1}{1+x^2} quickly?

Set y=\tan^{-1}(x), so \tan(y)=x . Differentiate: \sec^2(y)\,y'=1 . Then y'=\dfrac{1}{\sec^2(y)}=\cos^2(y) . Since \tan(y)=x , \cos^2(y)=\dfrac{1}{1+x^2} .

Q: What is \dfrac{d}{dx}\left(\left(\tan^{-1}(x)\right)^2\right) ?

Use chain rule: 2\tan^{-1}(x)\cdot \dfrac{1}{1+x^2}=\dfrac{2\tan^{-1}(x)}{1+x^2} .

Q: What is the derivative of \tan^{-1}\left(\frac{1}{x}\right) ?

Let u=\dfrac{1}{x}, so u'=-\dfrac{1}{x^2}. Then \dfrac{d}{dx}\left(\tan^{-1}\left(\frac{1}{x}\right)\right)=\dfrac{-\frac{1}{x^2}}{1+\frac{1}{x^2}}=\dfrac{-1}{x^2+1} .

Answer

\( \tan^{-1}(x) \) means \(\arctan(x)\). Let \(y=\arctan(x)\). Then \(\tan(y)=x\).

Differentiating implicitly:

\[
\sec^2(y)\, \frac{dy}{dx}=1
\]
So
\[
\frac{dy}{dx}=\frac{1}{\sec^2(y)}=\cos^2(y)
\]
Using \(y=\arctan(x)\), we have \(\tan(y)=x\), so \(\cos^2(y)=\frac{1}{1+x^2}\).

\[
\frac{d}{dx}\left(\tan^{-1}(x)\right)=\frac{1}{1+x^2}
\]

Detailed Explanation

We want to find the derivative of \( \tan^{-1}(x) \). Let’s go step by step.

Step 1: Identify the function clearly.

The function is

\[
y = \tan^{-1}(x)
\]

This means \(y\) is the arctangent of \(x\).

Step 2: Use the standard derivative formula.

A known result from calculus is:

\[
\frac{d}{dx}\left(\tan^{-1}(x)\right) = \frac{1}{1+x^2}
\]

So the derivative is the reciprocal of \(1+x^2\).

Step 3: State the final derivative.

\[
\frac{d}{dx}\left(\tan^{-1}(x)\right) = \frac{1}{1+x^2}
\]

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Calculus FAQ

What is the derivative of \( \tan^{-1}(x) \)?

\( \dfrac{d}{dx}\left(\tan^{-1}(x)\right)=\dfrac{1}{1+x^2} \).

What is the derivative of \( \tan^{-1}(u) \) where \(u\) depends on \(x\)?

Use the chain rule: \( \dfrac{d}{dx}\left(\tan^{-1}(u)\right)=\dfrac{u'}{1+u^2} \).

Find \( \dfrac{d}{dx}\left(\tan^{-1}(3x)\right) \).

Let \(u=3x\), so \(u'=3\). Then \( \dfrac{d}{dx}\left(\tan^{-1}(3x)\right)=\dfrac{3}{1+(3x)^2}=\dfrac{3}{1+9x^2} \).

Find \( \dfrac{d}{dx}\left(\tan^{-1}\left(\dfrac{x-1}{x+1}\right)\right) \).

Let \(u=\dfrac{x-1}{x+1}\). Then \(u'=\dfrac{2}{(x+1)^2}\). So derivative is \( \dfrac{u'}{1+u^2}=\dfrac{\frac{2}{(x+1)^2}}{1+\left(\frac{x-1}{x+1}\right)^2} \).

How do you derive \( \dfrac{d}{dx}\left(\tan^{-1}(x)\right)=\dfrac{1}{1+x^2} \) quickly?

Set \(y=\tan^{-1}(x)\), so \( \tan(y)=x \). Differentiate: \( \sec^2(y)\,y'=1 \). Then \( y'=\dfrac{1}{\sec^2(y)}=\cos^2(y) \). Since \( \tan(y)=x \), \( \cos^2(y)=\dfrac{1}{1+x^2} \).

What is \( \dfrac{d}{dx}\left(\left(\tan^{-1}(x)\right)^2\right) \)?

Use chain rule: \( 2\tan^{-1}(x)\cdot \dfrac{1}{1+x^2}=\dfrac{2\tan^{-1}(x)}{1+x^2} \).

What is the derivative of \( \tan^{-1}\left(\frac{1}{x}\right) \)?

Let \(u=\dfrac{1}{x}\), so \(u'=-\dfrac{1}{x^2}\). Then \( \dfrac{d}{dx}\left(\tan^{-1}\left(\frac{1}{x}\right)\right)=\dfrac{-\frac{1}{x^2}}{1+\frac{1}{x^2}}=\dfrac{-1}{x^2+1} \).

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